2(x-3)^2+6=14 eching

$$2 ( x - 3 ) ^ { 2 } + 6 = 14$$

x uchun yechish

$$x=5$$

$$x=1$$

Guruhlash orqali faktorlash qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-3\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}-6x+9\right)+6=14$$

2 ga x^{2}-6x+9 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}-12x+18+6=14$$

24 olish uchun 18 va 6'ni qo'shing.

$$2x^{2}-12x+24=14$$

Ikkala tarafdan 14 ni ayirish.

$$2x^{2}-12x+24-14=0$$

10 olish uchun 24 dan 14 ni ayirish.

$$2x^{2}-12x+10=0$$

Ikki tarafini 2 ga bo‘ling.

$$x^{2}-6x+5=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon x^{2}+ax+bx+5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-6 ab=1\times 5=5$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. Faqat bundan juftlik tizim yechimidir.

$$a=-5 b=-1$$

x^{2}-6x+5 ni \left(x^{2}-5x\right)+\left(-x+5\right) sifatida qaytadan yozish.

$$\left(x^{2}-5x\right)+\left(-x+5\right)$$

Birinchi guruhda x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$x\left(x-5\right)-\left(x-5\right)$$

Distributiv funktsiyasidan foydalangan holda x-5 umumiy terminini chiqaring.

$$\left(x-5\right)\left(x-1\right)$$

Tenglamani yechish uchun x-5=0 va x-1=0 ni yeching.

$$x=5 x=1$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-3\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}-6x+9\right)+6=14$$

2 ga x^{2}-6x+9 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}-12x+18+6=14$$

24 olish uchun 18 va 6'ni qo'shing.

$$2x^{2}-12x+24=14$$

Ikkala tarafdan 14 ni ayirish.

$$2x^{2}-12x+24-14=0$$

10 olish uchun 24 dan 14 ni ayirish.

$$2x^{2}-12x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -12 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 10}}{2\times 2}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 10}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 10}}{2\times 2}$$

-8 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2\times 2}$$

144 ni -80 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\times 2}$$

64 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±8}{2\times 2}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±8}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{12±8}{4}$$

x=\frac{12±8}{4} tenglamasini yeching, bunda ± musbat. 12 ni 8 ga qo'shish.

$$x=\frac{20}{4}$$

20 ni 4 ga bo'lish.

$$x=5$$

x=\frac{12±8}{4} tenglamasini yeching, bunda ± manfiy. 12 dan 8 ni ayirish.

$$x=\frac{4}{4}$$

4 ni 4 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=5 x=1$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-3\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(x^{2}-6x+9\right)+6=14$$

2 ga x^{2}-6x+9 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}-12x+18+6=14$$

24 olish uchun 18 va 6'ni qo'shing.

$$2x^{2}-12x+24=14$$

Ikkala tarafdan 24 ni ayirish.

$$2x^{2}-12x=14-24$$

-10 olish uchun 14 dan 24 ni ayirish.

$$2x^{2}-12x=-10$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-12x}{2}=-\frac{10}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{10}{2}$$

-12 ni 2 ga bo'lish.

$$x^{2}-6x=-\frac{10}{2}$$

-10 ni 2 ga bo'lish.

$$x^{2}-6x=-5$$

-6 ni bo‘lish, x shartining koeffitsienti, 2 ga -3 olish uchun. Keyin, -3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$x^{2}-6x+9=-5+9$$

-5 ni 9 ga qo'shish.

$$x^{2}-6x+9=4$$

x^{2}-6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-3\right)^{2}=4$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Qisqartirish.

$$x-3=2 x-3=-2$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$x=5 x=1$$